Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $r = \dfrac{8(2y - 1)}{y} \times \dfrac{y}{9(2y - 1)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ 8(2y - 1) \times y } { y \times 9(2y - 1) } $ $ r = \dfrac{8y(2y - 1)}{9y(2y - 1)} $ We can cancel the $2y - 1$ so long as $2y - 1 \neq 0$ Therefore $y \neq \dfrac{1}{2}$ $r = \dfrac{8y \cancel{(2y - 1})}{9y \cancel{(2y - 1)}} = \dfrac{8y}{9y} = \dfrac{8}{9} $